Answer
$\displaystyle \frac{32\ln 2}{3}-\frac{7}{4}$
Work Step by Step
$\left[\begin{array}{lll}
u=\ln x & e^{u}=x & x^{6}=e^{6u}\\
du=\frac{1}{x}dx && \\
x\in[1,2] & \Rightarrow u\in[0,\ln 2] &
\end{array}\right]$
$\displaystyle \int_{1}^{2}x^{5}\ln xdx=\int_{1}^{2}x^{6}\ln x\cdot\frac{1}{x}dx=\int_{0}^{\ln 2}ue^{6u}du$
by parts, $\displaystyle \left[\begin{array}{ll}
r=u & ds=e^{6u}du\\
dr=du & s=\frac{1}{6}e^{6u}
\end{array}\right],\quad\int_{0}^{\ln 2}rds=rs|_{0}^{\ln 2}-\int_{0}^{\ln 2}sdr$
$=\left[\dfrac{ue^{6u}}{6}\right]_{0}^{\ln 2}-\left[\dfrac{e^{6u}}{36}\right]_{0}^{\ln 2}$
$=\displaystyle \frac{2^{6}\cdot\ln 2}{6}-(\frac{2^{6}}{36}-\frac{1}{36})$
$=\displaystyle \frac{32\ln 2}{3}-\frac{63}{36}$
$=\displaystyle \frac{32\ln 2}{3}-\frac{7}{4}$
